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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /3161/CH9 | |
download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip |
initial commit / add all books
Diffstat (limited to '3161/CH9')
-rw-r--r-- | 3161/CH9/EX9.1/Ex9_1.sce | 33 | ||||
-rwxr-xr-x | 3161/CH9/EX9.2/Ex9_2.sce | 30 | ||||
-rw-r--r-- | 3161/CH9/EX9.3/Ex9_3.sce | 27 | ||||
-rwxr-xr-x | 3161/CH9/EX9.6/Ex9_6.sce | 43 |
4 files changed, 133 insertions, 0 deletions
diff --git a/3161/CH9/EX9.1/Ex9_1.sce b/3161/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..f2e27b7a3 --- /dev/null +++ b/3161/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,33 @@ +clc;
+//page 463
+//problem 9.1
+
+//Input signal strength Si = 0.5 W
+Si = 0.5;
+
+//Gaussian Power Spectral Density n = 10^(-10) W/Hz
+n = 10^(-10);
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Maximum frequency deviation Df = 60 kHz
+Df = 60 * 10^3;
+
+//Average power of the modulating signal mt = 0.1 W
+mt = 0.1;
+
+SNR = (3/(4*%pi^2))*((Df/fM)^2)*mt^2*(Si/(n*fM));
+
+disp('SNR is '+string(10*log10(SNR))+' dB');
+
+//Part b
+
+//Required SNR at output>40 dB = 10000
+
+//From (a), required Si/0.5 > 10000/4052.8
+//Or, required Si > 1.2337 W
+//Since, channel loss is 20 dB (=100),
+//Required transmitter power > 1.2337*100 = 123.37
+
+disp('Required transmitter power > 1.2337 x 100 = 123.37 ');
diff --git a/3161/CH9/EX9.2/Ex9_2.sce b/3161/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..7c1da149b --- /dev/null +++ b/3161/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,30 @@ +clc;
+//page 464
+//problem 9.2
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Maximum frequency deviation Df = 60 kHz
+Df = 60 * 10^3;
+
+//Figure of Merit for FM is G_FM
+G_FM = (3/2)*(Df/fM)^2;
+
+disp('Figure of Merit for FM system is '+string(G_FM));
+
+//Ratio of Figure of Merits of FM and AM systems is R
+R = G_FM/(1/3);
+
+disp('Ratio of Figure of Merits for FM and AM systems is '+string(R));
+
+Df_new = 2*Df;
+
+//Figure of Merit for FM when bandwidth is doubled is G_FM_new
+G_FM_new = (3/2)*(Df_new/fM)^2;
+
+//Ratio of Figure of Merits of FM and AM systems when bandwidth is doubled is R_new
+R_new = G_FM_new/(1/3);
+
+disp('Ratio of Figure of Merits for FM and AM systems when bandwidth is doubled is '+string(R_new));
+
diff --git a/3161/CH9/EX9.3/Ex9_3.sce b/3161/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..f3773ff67 --- /dev/null +++ b/3161/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,27 @@ +clc;
+//page 475
+//problem 9.3
+
+//Resistance R = 1000 Ohm
+R = 10^3;
+
+//Capacitance C = 0.1 * 10^-6 F
+C = 0.1*10^-6;
+
+//Break point for RC filter is f1
+f1 = 1/(2*%pi*R*C)
+
+//Baseband bandwidth of signal fM = 15 kHz
+fM = 15 * 10^3;
+
+Gain = atan(fM/f1)/(3*(f1/fM)*[1 - (f1/fM)*atan(fM/f1)]);
+
+disp('Initial Gain is '+string(10*log10(Gain))+' dB');
+
+//New Baseband bandwidth of signal fM_new = 15 kHz
+fM_new = 2*15 * 10^3;
+
+Gain_new = atan(fM_new/f1)/(3*(f1/fM_new)*[1 - (f1/fM_new)*atan(fM_new/f1)]);
+
+disp('Final Gain is '+string(10*log10(Gain_new))+' dB');
+
diff --git a/3161/CH9/EX9.6/Ex9_6.sce b/3161/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..6fcea73f4 --- /dev/null +++ b/3161/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,43 @@ +clc;
+//page 495
+//problem 9.6
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Carrier filter bandwidth is B = 60 kHz
+B = 60 * 10^3;
+
+//RMS frequency division Df_RMS = 30 kHz
+Df_RMS = 30 * 10^3;
+
+//Let a = Df_RMS/fM for substitution
+a = Df_RMS/fM;
+
+//Let b = fM/B for substitution
+b = fM/B;
+
+//Let input SNR 1 be I_SNR1 = 10 dB = 10
+I_SNR1 = 10;
+
+//Output SNR is O_SNR1
+O_SNR1 = (3*(a^2)*I_SNR1)/(1+6*((2/%pi)^0.5)*I_SNR1*exp(-(b)*I_SNR1));
+
+disp('Output SNR is '+string(10*log10(O_SNR1))+' dB');
+
+//Let input SNR 2 be I_SNR2 = 20 dB = 100
+I_SNR2 = 100;
+
+//Output SNR is O_SNR2
+O_SNR2 = (3*(a^2)*I_SNR2)/(1+6*((2/%pi)^0.5)*I_SNR2*exp(-(b)*I_SNR2));
+
+//Solution given in the book is 13.5431 which is fallacious, the correct answer is 24.32444
+disp('Output SNR is '+string(10*log10(O_SNR2))+' dB');
+
+//Let input SNR 3 be I_SNR3 = 30 dB = 1000
+I_SNR3 = 1000;
+
+//Output SNR is O_SNR3
+O_SNR3 = (3*(a^2)*I_SNR3)/(1+6*((2/%pi)^0.5)*I_SNR3*exp(-(b)*I_SNR3));
+
+disp('Output SNR is '+string(10*log10(O_SNR3))+' dB');
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